Question

A particle is projected with a velocity of 20 m/s at an angle of ${30}^{\circ }$ to an inclined plane of inclination ${30}^{\circ }$ to the horizontal. The particle hits the inclined plane at an angle ${30}^{\circ }$, during its journey. The time of flight in seconds is -

• A. $\frac{4}{\sqrt{3}}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $\frac{4}{\sqrt{3}}$

As we know that the time of flight
$T=\frac{2usin\theta }{g_{eff}}=\frac{2u_y}{g_{eff}}=\frac{2\times sin{30}^{\circ }\times 20}{gcos{30}^{\circ }}$
$T=\frac{40}{g\times \frac{\sqrt{3}}{2}}\frac{1}{2}=\frac{40}{\sqrt{3g}}=4/\sqrt{3}$.